1. ## Smooth

Let $\displaystyle f: \mathbb{R} \to \mathbb{R}$ be given. Assume that the square and cube of $\displaystyle f$ are smooth. Is $\displaystyle f$ smooth? That is , if $\displaystyle f \cdot f \in C^{\infty}$ and $\displaystyle f \cdot f \cdot f \in C^{\infty}$, does it follow that $\displaystyle f \in C^{\infty}$?

2. Or maybe we can put it this way: if $\displaystyle f^{n} \in C^{\infty}$ and $\displaystyle f^{n-1} \in C^{\infty}$ does it follow that $\displaystyle f \in C^{\infty}$?

3. Or maybe we can put it this way: if
$\displaystyle f^{n} \in C^{\infty}$ and $\displaystyle f^{n-1} \in C^{\infty}$ does it follow that $\displaystyle f \in C^{\infty}$ ?

Not in general. You need the extra assumption that $\displaystyle f^n,f^{n-1}$ are never zero.

The proof is easy: Choose any natural number $\displaystyle m$. Since $\displaystyle gcd(n,n-1)=1,$ there are integers $\displaystyle q, \ p$ such that $\displaystyle pn+q(n-1)=1$. So $\displaystyle m=pmn+qm(n-1)$; Thus $\displaystyle f^m=(f^n)^{mp}(f^{n-1})^{qm}$ is differentiable as a product of differentiable functions. Note that the exponents $\displaystyle mp, \ mq$ may be negative, and so the assumption $\displaystyle f^n, f^{n-1}\neq0$ is essential.

4. I have another way to show this.

If $\displaystyle f^3\not = 0$ for any point and $\displaystyle f^3\in C^{\infty}$ then $\displaystyle (f^3)^{1/3} \in C^{\infty}$ thus $\displaystyle f\in C^{\infty}$. Because the composition of infinitely differenciable functions is infinitely differenciable. Note, that $\displaystyle x^{1/3}$ is not differenciable at zero, which is why we need the condition that the function is never zero. However, I cannot come up with a conterexample.

This does not work with $\displaystyle f^2$.

5. Originally Posted by Rebesques
Not in general. You need the extra assumption that $\displaystyle f^n,f^{n-1}$ are never zero.

The proof is easy: Choose any natural number $\displaystyle m$. Since $\displaystyle gcd(n,n-1)=1,$ there are integers $\displaystyle q, \ p$ such that $\displaystyle pn+q(n-1)=1$. So $\displaystyle m=pmn+qm(n-1)$; Thus $\displaystyle f^m=(f^n)^{mp}(f^{n-1})^{qm}$ is differentiable as a product of differentiable functions. Note that the exponents $\displaystyle mp, \ mq$ may be negative, and so the assumption $\displaystyle f^n, f^{n-1}\neq0$ is essential.
Ah! So my thought that we may construct the function f as $\displaystyle \frac{f^3}{f^2}$ is at least not out of the ballpark. (This, of course, gives the same limitation that $\displaystyle f \neq 0$ at any point in its domain.) But this can't be the whole story because if we have
$\displaystyle f^2(x) = (x - 1)^4$
and
$\displaystyle f^3(x) = (x - 1)^6$

the function $\displaystyle f(x) = (x - 1)^2$ is, in fact, $\displaystyle C^{\infty}$.

How might we get around this? (Or am I barking up the wrong tree again? )

-Dan

6. Topsq, you have chosen a polynomial

Powers of that are always differentiable!

ps. What? 300 posts!! The Hacker must be feeling my breath on his back... :P

7. Originally Posted by Rebesques
Topsq, you have chosen a polynomial

Powers of that are always differentiable!

ps. What? 300 posts!! The Hacker must be feeling my breath on his back... :P
But my $\displaystyle f^3$ and $\displaystyle f^2$ are $\displaystyle C^{\infty}$ and there is a point in the domain of both that are 0. According to you and TPH my f should not be $\displaystyle C^{\infty}$. Or should we all be saying that we can't guarantee that such a $\displaystyle C^{\infty}$ f is possible under these conditions?

-Dan

8. According to you and TPH my f should not be...

Alright you win, add "polynomials and no-denominator ever functions out"